Prove by induction $\ n^{\log_2(n)} \le e^n $

Question

prove for any $\ n \ge 1 $

$\ n^{\log_2(n)} \le e^n $

Since both positive for every $\ n \ge 1 $ I was thinking of taking the $\ln $ of each function and then:

$\ln(n^{\log_2(n)}) \le \ln(e^n) \Rightarrow \log+2(n) \cdot \ln(n) \le n$

and for $\ n = 1 $ can see that $\log_2(1) \cdot 0 \le 1 $

the problem is I can't really do much with the function for $ n+1$

$\log_2(n+1) \cdot \ln(n+1) = ? $

can't really transform it to anything else.. I 've also tried taking the $\log_2 $ instead of $\ln $ but still..

Answer 1

Do you want to use induction only? Otherwise one can consider the function $f(x)= x-(lnx)^2$ in $x>1$ and see that the function is non-decreasing by using derivative test. Since $x>1$ and $f$ is non-decreasing, hence, $x-(lnx)^2>1>0$. Now take exponential both sides which will preserve the inequailty as exponential is increasing function and then you get the desired result.